A triangle has corners at (4 ,7 ), (1 ,4 ), and (6 ,2 ). What is the area of the triangle's circumscribed circle?

1 Answer
Dean R.
May 12, 2018

{841 pi}/98

Explanation:

I worked out the general form [here.](https://socratic.org/questions/a-triangle-has-corners-at-2-5-3-1-and-8-2-what-is-the-area-of-the-triangle-s-cir)

There's a formula we can just apply, but let's just work from the conclusion:

The squared radius of the circumcircle is the product of the squared sides of the triangle divided by 16(text{area})^2:

r^2 = {a^2b^2 c^2}/{ 16(text{area})^2 }

It's easier to leave everything squared. Since we need to generate the squared sides anyway, we note Archimedes' Theorem:

16(text{area})^2 = 4 a^2 b^2 - (c^2 -a ^2 - b^2)^2

r^2 = {a^2b^2 c^2}/{ 4 a^2 b^2 - (c^2 -a ^2 - b^2)^2 }

(4,7), (1,4), and (6,2)

We're free to assign a,b,c however we like. I prefer c to be the longest side.

a^2 = (4-1)^2 + (7-4)^2=18

b^2 = (4-6)^2+(7-2)^2=29

c^2 = (6-1)^2 + (2-4)^2 = 29

Oh, isosceles.

r^2 = {(18)(29)(29)}/{ 4 (18)(29) - (18^2) } = 841/98

The area of the circle is pi r^2 or {841 pi}/98

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